Making Math fun and accessible
If you’re going to submit a problem for MMM #13 and maybe win an awesome calculator from TI, you’d better hurry; they’re due in another 12 hours! When “time California” on Google says it’s 12:01 Tuesday morning, they’re due.
If you enjoyed this post, make sure you subscribe to my RSS feed!
Math can be fun. Really!
Check out this video.
If you enjoyed this post, make sure you subscribe to my RSS feed!
Blinkdagger and Wild About Math! are really stirring things up for Monday Math Madness (MMM) contests #13 and 14. And, Texas Instruments is giving away a VERY cool calculator (keep reading for more about the calculator), they’ll ship it internationally, plus we’ll allow anyone to win - even if you’ve won a prize in the MMM contest before. So, we expect submissions from everyone on the planet!
So, what are we doing differently? Well, we’re going to ask you guys and gals to submit your favorite MMM-caliber problems. That’s what you have to do be eligible to win MMM #13. We’ll pick the one we like best and then use it for MMM #14.
If you enjoyed this post, make sure you subscribe to my RSS feed!
It seems that Monday Math Madness #12 over at Blinkdagger is tougher than other MMM’s have been. So, there haven’t been many responses to date. So, your chances are higher than in previous contests of winning — if you can solve the problem!
If you enjoyed this post, make sure you subscribe to my RSS feed!
The prize for MMM #11 goes to Chao Xu. Here is Chao’s solution.
A number divisible by 11 if it’s alternating sum of it’s digits is divisible by 11.
We only have 6 numbers, so we have a+b+c-d-e-f divisible by 11, where a to f are variables for each position. The largest possible number formed by a+b+c-d-e-f is 6+5+4-3-2-1 = 9, the smallest number is -9,
Then we have only one number that is divisible by 11 and between 9 and -9 is 0.
We have
a+b+c = d+e+f
a+b+c = (a+b+c+d+e+f)/2
and a+b+c+d+e+f = 21.
a+b+c = 10.5
It’s not a whole number
It shows there is no number can be formed that’s divisible by 11.
If you enjoyed this post, make sure you subscribe to my RSS feed!
MMM #11 is a variation on MMM #9. I promise I won’t do any more variations on this problem after this one!
Consider all of the 6-digit numbers that one can construct using each of the digits between 1 and 6 inclusively exactly one time each. 123456 is such a number as is 346125. 112345 is not such a number since 1 is repeated and 6 is not used.
How many of these 6-digit numbers are divisible by 11?
While you may use a computer program to verify your answer, show how to solve the problem without use of a computer.
MMM #9 was interested in divisibility by 8. This contest is interested in divisibility by 11.
I have a Rubik’s Revolution, courtesy of Techno Source, (or $10 Amazon.com gift certificate) to give to the winner. I’ll give more than one prize if I get lots of correct submissions.
I’ve changed rule #9 to encourage original solutions, which I’m much more likely to acknowledge:
I may post names and website/blog links for people submitting timely correct well-explained solutions. I’m more likely to post your name if your solution is unique.
Here are the rules for the contest:
1. Email your answers with solutions to mondaymathmadness at gmail dot com.
2. Only one entry per person.
3. Each person may only win one prize per 12 month period. But, do submit your solutions even if you are not eligible.
3. Your answer must be explained. You must show your work! Wild About Math! and Blinkdagger will be the final judges on whether an answer was properly explained or not.
4. The deadline to submit answers is Tuesday, July 29, 2008 12:01AM, Pacific Time. (That’s Tuesday morning, not Tuesday night.) Do a Google search for “time California” to know what the current Pacific Time is.)
5. The winner will be chosen randomly from all timely well-explained and correct submissions, using a random number generator.
6. The winner will be announced Friday, August 1, 2008.
7. The winner (or winners) will receive a Rubik’s Revolution or a $10 gift certificate to Amazon.com. For those of you who don’t want a prize I’ll donate $10 to your favorite charity.
8. Comments for this post should only be used to clarify the problem. Please do not discuss ANY potential solutions.
9. I may post names and website/blog links for people submitting timely correct well-explained solutions. I’m more likely to post your name if your solution is unique.
If you enjoyed this post, make sure you subscribe to my RSS feed!
We have a winner for MMM #9. It’s Seb Perez-D. Congratulations! Drop me a line to arrange for your prize. And, check out Blinkdagger on Monday for their new problem.
A couple of people were confused about the deadline for submitting a solution. The deadline is 12:01 Tuesday Mountain Time, which is to say Tuesday morning. If you do a Google search for “time California” you’ll know what time it currently is in California. If it’s after 12:01AM the week after a contest is posted then you’re late! Using an offset from GMT will get you into trouble since California changes its offset twice each year.
If you enjoyed this post, make sure you subscribe to my RSS feed!
Last Friday Blinkdagger announced a winner for MMM #8. Here’s MMM #9:
Consider all of the 6-digit numbers that one can construct using each of the digits between 1 and 6 inclusively exactly one time each. 123456 is such a number as is 346125. 112345 is not such a number since 1 is repeated and 6 is not used.
How many of these 6-digit numbers are divisible by 8?
While you may use a computer program to verify your answer, show how to solve the problem without use of a computer.
If you enjoyed this post, make sure you subscribe to my RSS feed!
We have a winner for this seventh contest. Congratulations, Brent Yorgey! I’m delighted that Brent, of Math Less Traveled, won this one because Brent gives so much to students and readers of his blog. Brent - Enjoy your $25 gift certificate from our kind sponsor for this contest, the Art of Problem Solving.
Click here to see Brent’s solution, two of them actually.
If you enjoyed this post, make sure you subscribe to my RSS feed!
I’ve been wanting for some time to incorporate mathematical equations into my blog posts. What I’ve done up to now is to use this nice web site that Brent from Math Less Traveled pointed me to. You enter a LaTeX expression and the web-site creates an image which you then copy over to your web server and reference from your blog post. This system works great if you only need a small number of images. As an example, I used it to typeset the problem in Monday Math Madness #7.
If you enjoyed this post, make sure you subscribe to my RSS feed!
